Re: Capacitor and Force

On Fri, 12 Oct 2007 22:07:15 GMT, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:


"Tom Bruhns" <k7itm@xxxxxxx> wrote in message
news:1192219843.504190.120500@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Oct 12, 12:48 am, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx>
wrote:
Agreed. Getting the right idea about what's going on usually means
eliminating a lot of wrong - though superficially plausible - ideas.
Nobody talks much about the process of getting rid of the bad ideas,
but it always takes a while.

I think John Perry hit the nail on the head. I don't think anything is
wrong
with my analysis except I misunderstood how a ceramic capacitor worked.
I'm
going to try it with just two large plates and see what happens and if it
doens't work then I'll go back and look at my calculations. (I suppose I
should have tried that in the first places but large plates are harder to
manage)

Thanks,
Jon

As a check on your calculations, you might try calculating it in a
different way. For example, if you move a plate distance dx, and the
capacitance changes by dC in the process, and you still have the same
charge on the capacitor, you can (1) calculate the voltage change,
then (2) calculate the energy change, then (3) knowing the distance
and the energy change, calculate the force. Of course, the answer
should come out the same as your calculation concerning the force
attracting charges. Often, a good way to discover "wrong" ideas is to
try to arrive at the answer by as different a path as you reasonably
can.

well, all your doing is taking the derivatives then plugging them in. Its a
different path to another calculation that should work out too but if my
formulas are wrong in the first place or my logic is flawed then chances are
I would either get the right answer but still be wrong or get the wrong
answer and still be wrong ;)

I see what your saying though but it is a elementary problem and I seriously
doubt my calculations are wrong(I use maple to do them and I've done them
enough that any chance of it being wrong is pretty low). Now I might be
misapplying the formula but again, its pretty basic. If you have charges
then you have forces. Some thing that the distribution matters but it
doesn't. It is very similar to gravity(in fact it is equivalent except for
scale).

It isn't equivalent. For one thing, there is no such thing as shielding
with respect to gravitation. The presence of a sheet of lead between the
earth and moon would have no effect on the forces the earth exerts on the
moon; the forces exerted by the sheet of lead would be additive to the
earth's forces. But, if the moon and earth were oppositely charged, a
conducting shield between them would alter the electrostatic forces the
earth would exert on the moon.

Consider further that there is no such thing as a Faraday shield in
gravitation. In electrostatics, there is no electric field inside a hollow
conducting container. Even with a tiny hole in it to give access to the
inside, this is essentially true, but such an effect does not exist for
gravity. It's true that inside a hollow sphere made of some isotropic
material, there is no gravitational field, but this is only true for a
sphere. Inside a hollow non-spherical isotropic object, there is a
gravitational field. But in the world of electrostatics, it doesn't matter
what shape a hollow container has, there's no electric field inside. The
distribution of charge on the outside matters.

I'm sure they don't object to treating an object as a point mass at
the objects center of mass? Its always perfectly fine to do this(at least
with newtons laws it works out perfectly(can be proven)).


I believe John Perry solved the problem.


With regard to the force between charges which are not separated by
vacuum (or, essentially, by air), beware! Consider the following: if
I make an air-insulated capacitor with, say, 1000pF capacitance, and
put 1 nanocoulomb of charge on it, it will be charged to 1 volt. The
energy stored in it is C*(v^2)/2, or 0.5 nanojoule. If I replace the
air dielectric with a high-K ceramic, let's say with a relative
permittivity of 1000, then the capacitance is 1 microfarad. If I put
the same 1 nanocoloumb of charge on it, the voltage will be 1
millivolt: the capacitance went up 1000 times; the voltage went down
1000:1. Now what's the stored energy? That squared voltage term has
a bigger effect than the linear capacitance term: the stored energy
is 1/1000 as great as with the air dielectric, only half a picojoule.
THAT tells me that, since the spacing is the same, the FORCE is 1/1000
as much. So to the extent the ceramic "magnifies" the capacitance, to
the same extent, it reduces the force a given charge produces.


Yes, I know that. Its pretty simple since Q = C*V and in F = k*Q^2/r^2 =
k*(CV/r)^2. It means all factors(C,V,r) effect the force with the square.
This is sorta the problem and why such a large charge, or large voltage, or
very small distance must be used to get a significant force. Its not easy to
get them(if it was then life would be totally different).

Jon


.

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